Solutions to Problem Set 2

Math 310/520-01/02

9-8-2017

1. Construct a truth table for $(P
   \lor \lnot Q) \ifthen Q$ .

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & P & & Q & & $\lnot Q$ & & $P \lor \lnot Q$ & & $(P \lor \lnot Q) \ifthen Q$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & T & & F & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & F & & T & & T & & F & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & T & & F & & F & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & F & & T & & T & & F & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }}\quad\halmos $$


2. Construct a truth table for $(P
   \land \lnot Q) \lor (P \ifthen R)$ .

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & P & & Q & & R & & $\lnot Q$ & & $P \land \lnot Q$ & & $P \ifthen R$ & & $(P \land \lnot Q) \lor (P \ifthen R)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & T & & T & & F & & F & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & T & & F & & F & & F & & F & & F & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & F & & T & & T & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & T & & F & & F & & T & & T & & F & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & T & & T & & F & & F & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & T & & F & & F & & F & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & F & & T & & T & & F & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & F & & F & & F & & T & & F & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }}\quad\halmos $$


3. Use a truth table to prove that $(P \land Q) \ifthen P$ is a tautology. (This is called decomposing a conjunction.)

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & P & & Q & & $P \land Q$ & & $(P \land Q) \ifthen P$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & T & & T & & T & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & T & & F & & F & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & F & & T & & F & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & F & & F & & F & & T & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

Since $(P \land Q) \ifthen P$ is always true, it is a tautology.


4. In each case, determine whether the statement is true or false. Explain your answers.

(a) "If $\pi = 3$ , then Calvin Butterball got a parking ticket."

(b) "If $\sqrt{25} = 5$ , then $\sqrt{4} = -2$ ."

(a) The statement "$\pi =
   3$ " is false. Since the if-part of the conditional is false, the conditional is true.

(b) The statement "$\sqrt{25} = 5$ " is true, but the statement "$\sqrt{4} = -2$ " is false. Hence, the conditional is false.


5. For the conditional statement "If $x = y$ , then the sky is blue", write the converse, the inverse, and the contrapositive in words.

Converse: "If the sky is blue, then $x = y$ ".

Inverse: "If $x \ne y$ , then the sky isn't blue".

Contrapositive: "If the sky isn't blue, then $x \ne y$ ".


6. Use DeMorgan's Laws to negate each statement, then write the negation in words.

(a) "Calvin buys the stromboli or Phoebe does not eat the hamburger."

(b) "$\pi$ is not rational and $(\sin x)^2 + (\cos x)^2 = 53$ ."

(c) "If Bonzo has chicken pox, then the class will be dismissed."

(a) The negation is "Calvin does not buy the stromboli and Phoebe eats the hamburger".

(b) The negation is "$\pi$ is rational or $(\sin x)^2 + (\cos x)^2 \ne 53$ ".

(c) By conditional disjuntion, the original statement is equivalent to "Bonzo does not have chicken pox or the class will be dismissed".

The negation is "Bonzo has chicken pox and the class will not be dismissed".


[Math 520]

7. Use a truth table to show that the following statements are logically equivalent:

$$\lnot(P \iff Q) \quad\hbox{and}\quad (P \land \lnot Q) \lor (Q \land \lnot P).$$

$$\matrix{ P & Q & \lnot P & \lnot Q & P \land \lnot Q & Q \land \lnot P & (P \land \lnot Q) \lor (Q \land \lnot P) \cr \noalign{\vskip2pt \hrule \vskip2pt} \cr T & T & F & F & F & F & F \cr T & F & F & T & T & F & T \cr F & T & T & F & F & T & T \cr F & F & T & T & F & F & F \cr }$$

$$\matrix{ P & Q & P \iff Q & \lnot (P \iff Q) \cr \noalign{\vskip2pt \hrule \vskip2pt} \cr T & T & T & F \cr T & F & F & T \cr F & T & F & T \cr F & F & T & F \cr }$$

The columns for $\lnot(P \iff
   Q)$ and $(P \land \lnot Q) \lor (Q \land \lnot
   P)$ are the same, so the statements are logically equivalent.



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